How Do Charges Distribute in Series-Connected Capacitors?

[i_island0]

Applying Kirchoff's 2nd law we have the ODE

q'(t)+\frac{q(t)}{\tau}=\frac{V}{R}-\frac{q_o}{\tau},\quad q(0)=0 with \tau=R\,C, \quad \frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}

:

don't you think that the differential equation would be:
(dq/dt) R + (q + qo)/C1 + q/C2 - V = 0

 

[dynamicsolo]

For start let the maths talk.
The charge of every capacitor is symbolized by q_1(t),\,q_2(t) and the current i=\frac{d\,q}{d\,t} where q is the charge "running" in the circuit.
...Thus at every instant we have

q_1(t)=q_o+q, \quad q_2(t)=q \quad (1)
...

Thus, never mix the charge of the capacitor and the charge "running" in the circuit

I suppose if you start with such a premise, you can arrive at such a conclusion. But there still seems to be resistance to answering a question I already raised: why would a charge placed on one capacitor, made of conductive material and directly connected by conductive wire to a second capacitor, also made of conductive material, have no influence on the charges in the second capacitor? How could applying charge to C1 not produce charge separation in C2 when the two capacitors are connected even before the switch is closed? That situation is unstable and C2 would not be starting out uncharged before the switch is closed.

But this still doesn't matter to the charge configuration after the switch is closed and the capacitors reach equilibrium with the battery. There still seems to be an insistence that somehow the applied charge q0 "knows" it is applied charge and doesn't count toward what is happening on the plates in the capacitors. What is the physical explanation for counting it separately from whatever else is happening on the plates in the two capacitors? This description of the capacitor charges

q_1(t)=q_o+q, \quad q_2(t)=q \quad

treats q0 as if it somehow has nothing to do with what is otherwise happening in the circuit, so it is not surprising that it just rides out the charging of the capacitors and so naturally ends up in the mathematical solution. (Of course the charges on the two capacitors end up different if you claim that.) C1 seems to have been permanently set to have a charge q0 greater than the charge on C2 and nothing appears able to change that, which seems to have no physical justification...

 

[i_island0]

i checked the equilibrium condition using Gauss law.. and according to my calculation it is in equilibrium already, do you mind checking that too.

 

[Shooting Star]

But there still seems to be resistance to answering a question I already raised: why would a charge placed on one capacitor, made of conductive material and directly connected by conductive wire to a second capacitor, also made of conductive material, have no influence on the charges in the second capacitor? How could applying charge to C1 not produce charge separation in C2 when the two capacitors are connected even before the switch is closed? That situation is unstable and C2 would not be starting out uncharged before the switch is closed.

Hi dynamicsolo,

This was the situation I was interested in. I wanted to know what charges on each capacitor would be before the switch was closed. In fact, I just want to know what happens when C1 has q0 and -q0, and is connected to C2 which is uncharged. This is different from the problem that the OP had posted.

 

[Rainbow Child]

I suppose if you start with such a premise, you can arrive at such a conclusion. But there still seems to be resistance to answering a question I already raised: why would a charge placed on one capacitor, made of conductive material and directly connected by conductive wire to a second capacitor, also made of conductive material, have no influence on the charges in the second capacitor? How could applying charge to C1 not produce charge separation in C2 when the two capacitors are connected even before the switch is closed? That situation is unstable and C2 would not be starting out uncharged before the switch is closed.

Due to the length of my post you may not have noticed:

...
\hline

^1 To set the problem in the correct basis, we must say that we do two things simultaneously. We close the switch S and connect the two capacitors at the same time. If not we have the "new" problem. It needs a post of it's own

But this still doesn't matter to the charge configuration after the switch is closed and the capacitors reach equilibrium with the battery. There still seems to be an insistence that somehow the applied charge q0 "knows" it is applied charge and doesn't count toward what is happening on the plates in the capacitors. What is the physical explanation for counting it separately from whatever else is happening on the plates in the two capacitors? ...

The physical expanation is quite simple. From the definition of the capacity and the definition of the current.

1. Capacity: C=\frac{q}{V} The charge q is the absolute value of the charge "sitting" on of the plates of the capacitor.
2. Current: i=\frac{d\,q}{d\,t} The charge q represents the charge d\,q, passing through the cross-section of the wire at the time interval d\,t.

This description of the capacitor charges

q_1(t)=q_o+q, \quad q_2(t)=q \quad

treats q0 as if it somehow has nothing to do with what is otherwise happening in the circuit, so it is not surprising that it just rides out the charging of the capacitors and so naturally ends up in the mathematical solution. (Of course the charges on the two capacitors end up different if you claim that.) C1 seems to have been permanently set to have a charge q0 greater than the charge on C2 and nothing appears able to change that, which seems to have no physical justification...

Who said that q_o is permanent ?
To make things more clear in order to explain the physical interpretation of q_o consider the following example.

Consider a capacitor, with capacity C whih is connected through a resistance R with an ideal battery of voltage V and let it there until it is full charged. The final charge on the capacitor is q_o=C\,V.
Now disconnect the capacitor and set up a new circuit with the same elements, but now connect the negative plate (say A) with the positive pole of the battery, and the positive plate (say B)with the negative pole of the battery. Which is now the function Q_A(t) describing the charge on the capacitor's plate A?
I think you agree that firstly the capacitor will discharged and then will be charged up to voltage V. Correct?
If you do agree, that means that the final charge on the plate A will be Q_A(\infty)=C\,V, while the final (=total) charge moved by the battery is q(\infty)=2\,C\,V. Correct?
Now if you don't distinguish between the charge q,\,q(0)=0 running through the circuit and the chagre Q_A(t),\,Q_A(0)=-q_o=-C\,V you will run into problems.

The charge q is given by

q(t)=2\,C\,V\,(1-e^{-t/\tau}), \tau=R\,C

while the charge Q_A on the plate is

Q_A(t)=C\,V\,(1-2\,e^{-t/\tau}), \tau=R\,C

Does this clears out the situation?

 

[i_island0]

ok.. your explanation works well with what i have read